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Evgesh-ka [11]
4 years ago
12

tyrosine kinase inhibitor binds and inhibits BTK. As a result of the experiment, you are able to elute BTK from the column, but

in a mixture of other tyrosine kinases. Why are tyrosine kinases other than BTK present in the eluate?
Chemistry
1 answer:
Sindrei [870]4 years ago
3 0

Answer: It is because tyrosine kinases and BTK have similar solubilities

Explanation:

In column chromatography, components of a mixture are seperated based on their relative solubilities in two non-mixing phases.

In essence, tyrosine kinases and BTK are present in the eluate due to their similar solubility rates that arise from the similar chemical structure both possess (otherwise it would be impossible for the inhibitor meant for Tyrosine kinase to bind and also inhibits BTK)

Thus, the similar solubilities of both groups is the reason they could elute out of the column without being adsorped.

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The Battle of Gettysburg proved to be the turning point of the war.

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3 years ago
Using this balanced equation: 2 NaOH + H2SO4 —> H2O + Na2SO4
mars1129 [50]
<h3>Answer:</h3>

266.325 g

<h3>Explanation:</h3>

We are given the balanced equation;

2NaOH + H₂SO₄ → H₂O + Na₂SO₄

  • Mass of NaOH as 150 g

We are required to determine the mass of Na₂SO₄ that will be formed.

<h3>Step 1: Determine the number of moles of NaOH</h3>

Moles = Mass ÷ molar mass

Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 150 g ÷ 40 g/mol

                          = 3.75 moles

<h3>Step 2: Determine the number of moles of sodium sulfate formed</h3>
  • From the equation 2 moles of NaOH reacts with sulfuric acid to form 1 mole of sodium sulfate.
  • Therefore; mole ratio of NaOH : Na₂SO₄ is 2 : 1

Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2

                                      = 3.75 moles ÷ 2

                                     = 1.875 moles

<h3>Step 3: Determine the mass of Na₂SO₄ produced.</h3>

we know that;

Mass = Moles × Molar mass

Molar mass of Na₂SO₄ is 142.04 g/mol

Therefore;

Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol

                           = 266.325 g

Thus, the mass of sodium sulfate formed 266.325 g

7 0
4 years ago
A certain crunch cereal contains 11.0 grams of sugar(sucrose, C12H22O11) per serving size of 60.0 gram. How many
Paladinen [302]

Answer:

51.859 grams

Explanation:

Given:

Mass of sugar in the crunch cereal per serving= 11.0 grams

Mass of cereal served per serving = 60.0 grams

Molar mass of the sugar (C₁₂H₂₂O₁₁) = ( 12 × 12 + 22 × 1 + 16 × 11 ) = 342 grams

Number of moles of sugar per serving = Mass / Molar mass

= 11.0 / 342

= 0.03216 moles

Now,

0.03216 moles sugar requires 60 grams cereal

for 0.0278 moles of sugar, cereal required = (60 / 0.03216) × 0.0278

or

for 0.0278 moles of sugar, cereal required = 51.859 grams

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